# Solving $y''+2y'+y = 2e^{-t}$ by the method of undetermined coefficients

I need to solve

$$y''+2'y+y = 2e^{-t},$$

using the method of undetermined coefficients (and by founding a solution for the homogeneous equation).

I tried first guessing a solution of the form $y = Ae^{-t}$, but when I tried to solve for $A$, I got a surprise: I couldn't equate the terms when I plugged $y$ in the differential equation, because I got $e^{-t}(A-2A+2A) = 2e^{-t}$ but the $A$'s sum to $0$. I then searched my book and realized that the problem was because $Ae^{-t}$ is already a solution for the homogeneous equation. I then tried $Ate^{-t}$ because my book tries it for a different equation and it worked, but in my case I got the $A$'s summing to $0$ again.

What should be my guess?

The other answers did not address why one can use the auxiliary equation to find the complementary solution.

$$\begin{array}{rcl} y'' + 2y' + y &=& 0 \\ y'' + y' + y' + y &=& 0 \\ e^t(y'' + y') + e^t(y' + y) &=& 0 \\ (e^t y')' + (e^t y)' &=& 0 \\ e^t y' + e^t y &=& A \\ (e^t y)' &=& A \\ e^t y &=& A t + B \\ y &=& A t e^{-t} + B e^{-t} \\ \end{array}$$

Using the same method, one doesn't actually need to use the method of undetermined coefficients:

$$\begin{array}{rcl} y'' + 2y' + y &=& 2e^{-t} \\ y'' + y' + y' + y &=& 2e^{-t} \\ e^t(y'' + y') + e^t(y' + y) &=& 2 \\ (e^t y')' + (e^t y)' &=& 2 \\ e^t y' + e^t y &=& 2t + A \\ (e^t y)' &=& 2t + A \\ e^t y &=& t^2 + A t + B \\ y &=& t^2 e^{-t} + A t e^{-t} + B e^{-t} \\ \end{array}$$

You could never have guessed the particular solution $y=t^2e^{-t}$ using the undetermined coefficients method.

Verification:

$$\begin{array}{rcl} y &=& t^2 e^{-t} + A t e^{-t} + B e^{-t} \\ y' &=& (2t - t^2) e^{-t} + A (1 - t) e^{-t} - B e^{-t} \\ &=& - t^2 e^{-t} + (2 - A) t e^{-t} + (A - B) e^{-t} \\ y'' &=& - (2t - t^2) e^{-t} + (2 - A) (1 - t) e^{-t} - (A - B) e^{-t} \\ &=& t^2 e^{-t} + (A - 4) t e^{-t} + (2 + B - 2A) e^{-t} \\ y'' + 2y' + y &=& [1 + 2(-1) + 1] t^2 e^{-t} + [(A - 4) + 2(2 - A) + A] t e^{-t} + [(2 + B - 2A) + 2(A - B) + B] e^{-t} \\ &=& 2 e^{-t} \end{array}$$

Conclusion... don't use the method of undetermined coefficients?

• What you mean "you couldn't have guessed the particular solution"? The two homogeneous solutions are $e^{-t}$ and $te^{-t}$, so common sense would indicate to try $At^2e^{-t}$. Saying not to use undetermined coefficients is bad advice. While your method is nice, it's not what the question is asking. – Dylan Dec 31 '17 at 7:41

Homogeneous equation:

$y''+2y'+y=0$, $x^2+2x+1=(x+1)^2$ solution $ae^{-t}$

$y(t)=a(t)e^{-t}$, $y'(t)=a'(t)e^{-t}-a(t)e^{-t}$, $y''=a''(t)e^{-t}-a'(t)e^{-t}-a'(t)e^{-t}+a(t)e^{-t}$,

$y''2y'+y=a"(t)e^{-t}=2e^{-t}$, $a''(t)=2, a(t)=t^2+at+b, y(t)=(t^2+at+b)e^{-t}$.

• Sorry, I didn't understand what happened in the last line – Guerlando OCs Oct 23 '16 at 21:03
• I just compute $y"+2y'+y$ which is equal to $a"(t)e^{-t}$ then I solve $a"(t)e^{-t}=2e^{-t}$ this implies $a(t)=t^2+2a+b, y=a(t)e^{-t}=(t^2+2a+b)e^{-t})$. – Tsemo Aristide Oct 23 '16 at 21:06
• So I should always try $p(t)\cdot solution$? – Guerlando OCs Oct 23 '16 at 21:16
• Also, in this example, $a'(t)$ and $a(t)$ canceled. Will it always be the case? I think it'd be harder to solve if they didn't – Guerlando OCs Oct 23 '16 at 21:23

Some of the answers here beat around the bush. You're supposed to multiply the particular solution by x raised to the multiplicity of "b" in a form $e^{bt}$. -1 is a double root with a multiplicity of 2 so your potential solution is multiplied by $x^2$.

In other words, the form of the particular solution is $x^2(A)e^{-t}$

I apologize for rehasing information but these answers beat around the bush. They make the method sound trivial like guesswork. In reality, the particular solution is a very specific formula. Also, your A's dont sum to 0, but you get junk nonetheless.

Hint: Multiply both sides by $e^t$ and then set $z = ye^t$ and express the equation in terms of $z$.

• The book didn't teach this so I don't know if my teacher would accept. Isn't there a rule to guess another type of solution? Because my book talked about this problem, when the guess is the solution of the homogeneous equations, then he tried the derivative of the guess, but didn't explain why. That's what I tried here. – Guerlando OCs Oct 23 '16 at 21:01
• @GuerlandoOCs i think this is variation by parameters. – The Great Duck Dec 21 '16 at 3:03

It happened because $te^{-t}$ is also a solution of the homogeneous equation. Try $At^2e^{-t}$ instead.

• $e^t$ or $e^{-t}$? – Guerlando OCs Oct 23 '16 at 21:02
• Sorry. That was a typo. I will edit now. – Prince Kumar Oct 23 '16 at 21:03
• But WHY is it a solution. Be specific! – The Great Duck Dec 21 '16 at 3:03

We must see the complementary solution before assuming the particular solution, and the reason is to prevent similarities which may occur between the two. For the question, the complementary solution is found when the D.E is homogeneous $$y''+2y'+y=0$$ $$r^2+2r+1=0$$ $$r_{1,2}=-1$$ so, the complementary solution will be $$y_c=c_1e^{-t}+c_2e^{-t}$$ to prevent the similarity between them, we should multiply one of them by $t$, so the solution will become $$y_c=c_1e^{-t}+c_2te^{-t}$$ the next step is to find the particular solution depending on the R.H.S which was $(e^{-t})$ If we assume the particular solution as $y_p=Ae^{-t}$, we will find this solution is found in complementary solution therefore we should multiply by $t$. The last thing is not enough because the solution $te^{-t}$ is also found in the complementary solution, so we should multiply by $t^2$

hence the assumed particular solution must be $$y_p=At^2e^{-t}$$

• should I always proceed this way? Multiplying bu $t$, and if this won't work, multiply by $t^2$? – Guerlando OCs Oct 23 '16 at 21:58
• yes, The main point is to prevent the similarity between the complementary solution or particular solution or between us. This occurs through multiplying by $t$ and then compare whether similarity if there are similarities, the power of $t$ must be increased – E.H.E Oct 23 '16 at 22:07